Delete positional parameters in Bash?

Question

You can skip positional parameters with shift but can you delete positional parameters by passing the position?

x(){ CODE; echo \"$@\"; }; x 1 2         


        

Answer 1

while loop over "$@" with shift + set: move each parameter from first to last position, except "test"

# remove option "test" from positional parameters
i=1
while [ $i -le $# ]
  do
    var="$1"
    case "$var" in
      test)
        echo "param \"$var\" deleted"
        i=$(($i-1))
      ;;
      *)
        set -- "$@" "$var"
      ;;
    esac
    shift
    i=$(($i+1))
done